A geometry problem involving triangles In the figure, AE is the bisector of the exterior angle CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, then find the length of CE.

My Attempt: I tried to find out the existence of congruent triangles in the diagram, but couldn't find any.
Any help will be appreciated.
 A: Hint : The cosine law of triangles will definitely solve the problem but that may lead to boring calculations. 
You can construct congruent triangles. 
Draw a line passing point E and parallel to AC, intersecting AB at F. Now ABC is congruent to FBE, and AF=EF, so you can find the length of AF by solving the equation w.r.t AF.

$\frac{BA}{BA+AF}=\frac{AC}{EF}=\frac{AC}{AF}$

Continue the reasoning and you can find the length of CE. 
A: Let $CE=x$.
Hence, by bisector theorem we obtain:
$$\frac{AB}{AC}=\frac{BE}{EC}$$ or
$$\frac{10}{6}=\frac{x+12}{x}$$ or
$$\frac{5}{3}=1+\frac{12}{x}$$ or
$$\frac{12}{x}=\frac{2}{3}$$ or
$$x=18$$.
A: Alternatively: Continue the line $D$ until $F$ and connect $F$ with $E$ so that $AC=AD$.
Note that $\Delta ACE$ is equal to $\Delta ADE$, because two sides and the angle between them are equal. It implies the line $AE$ is a bisector in $\Delta BDE$.
Using the property of bisector:
$$\frac{AB}{BE}=\frac{AD}{DE} \Rightarrow \frac{10}{12+CE}=\frac{6}{CE} \Rightarrow CE=18.$$
